LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every ordered field is an ordered ring, and its order is the one its positive cone induces

Statement

Let $F$ be an ordered field with positive cone $P$ (Ordered field), and let $\leq$ be the relation $a \leq b : ⟺ (b - a \in P$ or $a = b)$ that Ordered field defines from $P$ . Then:

$F$ with the operations of Field is a commutative ring (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring), and $P$ is a cone in the sense of The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order;
$\leq$ is a total order making $F$ an ordered ring (Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication), whose positive cone ${x \in F : 0 < x}$ is exactly $P$ ;
$1 \in P$ , that is $0 < 1$ .

So an ordered field is an ordered ring, and its order and its positive cone determine each other exactly as they do in any ordered ring.

Facts & Assumptions

Given: An ordered field $F$ with positive cone $P$ , and $\leq$ defined from $P$ by $a \leq b : ⟺ (b - a \in P$ or $a = b)$ (Ordered field).

[A1]

Axiom (O1): for each $x \in F$ exactly one of $x \in P$ , $x = 0$ , $- x \in P$ holds (Ordered field).

[A2]

Axiom (O2): if $x, y \in P$ then $x + y \in P$ and $x y \in P$ (Ordered field).

[A3]

The order of an ordered field is defined by $a < b : ⟺ b - a \in P$ , and $a \leq b$ means $a < b$ or $a = b$ (Ordered field).

[L1]

Every field is a commutative ring with $1 \neq 0$ (Every field is a commutative ring with $1 \neq 0$ ; it is an integral domain, and it is a commutative division ring, Field).

[L2]

In a ring, a subset satisfying trichotomy and closure under addition and multiplication is a cone, and the relation it induces is a total order making the ring an ordered ring whose positive cone is that subset (The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order, Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication).

[L3]

In an ordered field, $a \neq 0$ implies $a^{2} > 0$ , that is $a \cdot a \in P$ (Squares of nonzero elements are positive).

Proof

technique · direct

1.1

$F$ is a commutative ring under its own addition and multiplication, with the same $0$ and $1$ .

2.1

$P$ is a cone in the ring $F$ : trichotomy is axiom (O1) verbatim, and closure under addition and under multiplication is axiom (O2) verbatim. This proves claim 1.

step 1.1A1A2L2

3.1

By [L2] applied to the ring $F$ and the cone $P$ , the relation $a \leq_{P} b : ⟺ (b - a \in P$ or $a = b)$ is a total order making $F$ an ordered ring, and its positive cone is $P$ .

step 2.1L2

4.1

That relation is the order of the ordered field: [A3] defines $a < b$ as $b - a \in P$ and $a \leq b$ as $a < b$ or $a = b$ , which is the definition of $\leq_{P}$ word for word. So $\leq$ and $\leq_{P}$ are the same relation, and claim 2 follows.

step 3.1A3

5.1

Claim 3: $1 \neq 0$ by [L1], so $1 \cdot 1 \in P$ by [L3]; and $1 \cdot 1 = 1$ because $1$ is the multiplicative identity. Hence $1 \in P$ , that is $0 < 1$ .

step 4.1L1L3∎

Remarks

This is one of the four bridges the page carries. The published Ordered field and the Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication of this page are two presentations of compatible orders, and without this item a later result about ordered rings could not be applied to $Q$ or $R$ without redoing the translation each time.
The converse is false as stated, and no converse is claimed. $Z$ is an ordered ring and is not a field at all, as the companion page records. What is true, and is claim 2, is that on a field the two presentations of an order agree.
Where the strict form of (OR2) is used. Axiom (O2) of Ordered field is already strict, so claim 1 is a verbatim match with the cone conditions of The order presentation and the positive-cone presentation of an ordered ring determine each other: $P = {x : 0 < x}$ satisfies trichotomy and closure, and $a < b : ⟺ b - a \in P$ recovers the order and no strengthening is needed; that is exactly why Ordered ring: a ring with a total order compatible with addition and with positives closed under multiplication states its multiplicative axiom in the strict form.

Depends on

Used by

ℚ and ℝ are fields, hence commutative rings, integral domains and ordered rings, all of characteristic 0 Example

Dependency tree · two levels

21 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ordered field (Wikipedia) (standard reference, not scraped)
Ordered ring (Wikipedia) (standard reference, not scraped)
Field (mathematics) (Wikipedia) (standard reference, not scraped)